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Sequentially compact space

  1. Mar 16, 2010 #1
    Hello Physicsforums!
    I have a problem with the difference between complete metric space and a sequentially compact metric space.
    For the first one every Cauchy sequence converges inside the space, which is no problem.
    But for the last one "every sequence has a convergent subsequence." (-Wiki) And it's here that I get lost.

    How does this affect the constraints on the space?
    Could someone please try to give me an intuitive explanation?

    For [1,9] on the real axis we can take the sequence (1,2,3,4,5,6) as an example. How do we find a convergent subsequence in this one?
    Have I missunderstood it all?
     
  2. jcsd
  3. Mar 16, 2010 #2

    George Jones

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    What is the definition of "sequence"?
     
  4. Mar 16, 2010 #3
    Hmm, I use this one http://en.wikipedia.org/wiki/Sequence .
    With
    Maybe it´s here that I am confused. :uhh:

    Should we only work with Cauchy sequences maybe?
     
  5. Mar 16, 2010 #4
    These definitions apply to infinite sequences. (1,2,3,4,5,6) is not an infinite sequence. It doesn't even mean anything for a finite sequence to converge!
     
  6. Mar 16, 2010 #5

    jav

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    To the original question..

    In a complete metric space (an) converges <-> (an) is cauchy

    In a compact metric space, every sequence an contains a convergent subsequence (ank).

    We should note that convergence -> cauchy in any metric space.

    Then, in a compact metric space, every sequence an contains a cauchy subsequence (ank).

    Regardless, the properties of these two types of spaces are completely different.

    A simple example highlighting the difference between the two is a subset of R1. Consider, the interval (0,1).

    By the Heine-Borel theorem, this space is not compact since it is not closed.

    It is, however, a complete metric space since cauchy <-> convergent in R1.

    Was this your question?
     
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